Paper 2 Section B Practice: Hypothesis Testing — Answers

Questions: Hypothesis Testing Drills.

Question 1

(a)

$$\bar{x}=\frac{17859.6}{36}=496.1\text{ ml}.$$

(b) Let $\mu$ be the population mean amount of liquid in a bottle, in ml.

$$H_0:\mu =500,H_1:\mu <500.$$

(c) Under $H_0$,

$$\bar{X}\sim N\left(500,\frac{12^2}{36}\right)$$

exactly if the population is normal, or approximately if a CLT approximation is appropriate. The observed statistic is

$$z=\frac{496.1-500}{12/\sqrt{36}}=-\mathrm{1.95.}$$

For a lower-tailed $5\%$ test, the critical value is approximately $-1.645$. Since

$$-1.95<-1.645,$$

reject $H_0$.

(d) There is sufficient evidence at the $5\%$ significance level that the population mean fill amount is less than $500$ ml. This concerns the population mean; individual bottle amounts vary, so the conclusion does not imply that every bottle contains less than $500$ ml.

Question 2

(a) The phrase “has changed” allows departures both below and above $48$ hours, so $H_1$ must be two-tailed.

(b) Let $\mu$ be the population mean battery lifetime, in hours.

$$H_0:\mu =48,H_1:\mu \ne 48.$$

(c) Under $H_0$,

$$\bar{X}\sim N\left(48,\frac{5.4^2}{64}\right).$$

The observed statistic is

$$z=\frac{46.7-48}{5.4/\sqrt{64}}=-1.925925\dots .$$

For a two-tailed $1\%$ test, the critical values are approximately $\pm 2.576$. Since

$$|z|<2.576,$$

do not reject $H_0$. There is insufficient evidence at the $1\%$ significance level that the population mean battery lifetime differs from $48$ hours.

(d)

$$p=2P(Z<-1.925925\dots )=0.0541136\dots \approx 0.0541$$

to 3 significant figures. Since $0.0541>0.01$, the p-value method gives the same decision.

Question 3

(a) Let $\mu$ be the population mean score after participation in the training programme.

$$H_0:\mu =72,H_1:\mu >72.$$

(b)

$$z=\frac{74.4-72}{8/\sqrt{49}}=\mathrm{2.10.}$$

(c) The upper $5\%$ critical region begins at $z=1.645$. Since $2.10>1.645$, reject $H_0$.

There is sufficient evidence at the $5\%$ significance level that the population mean score after participation in the programme exceeds $72$. This observational comparison alone does not establish that the programme caused the increase.

(d) Because the population standard deviation is known, the standard error under $H_0$ is

$$\frac{8}{\sqrt{49}}.$$

It supplies the scale used to standardise $\bar{X}$ and permits the core known-variance $Z$-test.

(e) One suitable condition is that the observations form a random independent sample. Alternatively, the population is normal or the sample size is sufficiently large for the distribution of $\bar{X}$ to be approximately normal.

Question 4

(a) For a lower-tailed $5\%$ test, the critical value is approximately

$$-\mathrm{1.645.}$$

(b) The standard error is

$$\frac{10}{\sqrt{25}}=2.$$

Reject $H_0$ when

$$\frac{\bar{x}-120}{2}<-1.644853\dots .$$

Therefore

$$\bar{x}<116.710292\dots .$$

The largest sample mean to 2 decimal places that lies in the rejection region is

$$116.71\text{ seconds}.$$

(c) If $\bar{x}=117.0$,

$$z=\frac{117.0-120}{2}=-\mathrm{1.50.}$$

Since $-1.50>-1.645$, do not reject $H_0$.

(d) The rejection boundary for the sample mean is

$${\mu }_0+z_{0.05}\frac{\sigma }{\sqrt{n}}.$$

As $n$ increases, $\sigma /\sqrt{n}$ decreases, so the boundary moves closer to ${\mu }_0=120$. The significance level fixes the standard-normal critical value, but the reduced variability of $\bar{X}$ means that a smaller departure from $120$ can be statistically significant.